The increasing order of the following compounds towards HCN addition is
(i) 4-methoxybenzaldehyde
(i

Q: Why does the methoxy group ($-OCH_3$) act as an electron donor in compound (i) but an electron withdrawer in compound (iv)?

It all comes down to positioning! In compound (i), the $-OCH_3$ is at the para position, allowing its lone pairs to participate in resonance ($+R$ effect), which overpowers its weak inductive pull ($-I$). In compound (iv), it is at the meta position. Resonance cannot operate from the meta position to the carbonyl group, so we only observe its electron-withdrawing inductive effect ($-I$).

Q: Why is p-nitrobenzaldehyde more reactive than benzaldehyde?

The $-NO_2$ group is a strong Electron Withdrawing Group (EWG). Through both its $-I$ and $-R$ effects, it pulls electron density away from the benzene ring and the attached carbonyl carbon. This increases the positive charge on the carbonyl carbon, making it a much more attractive target for the incoming $CN^-$ nucleophile.

Q: Does the inductive effect operate from all positions?

Yes, the inductive effect operates through sigma bonds and is felt from ortho, meta, and para positions. However, it is distance-dependent, meaning it is strongest at the ortho position and weakest at the para position.

Q: What would happen if we had an $-OH$ group instead of an $-OCH_3$ group?

The $-OH$ group behaves very similarly to the $-OCH_3$ group. It has a strong $+R$ effect from the ortho and para positions (making the carbonyl less reactive) and a $-I$ effect from the meta position (making the carbonyl slightly more reactive).

Q: Why do we even care about the $\delta+$ charge on the carbonyl carbon?

Because the first step of this reaction is the attack of a nucleophile (a species that loves positive charge). The more positive the target carbon is, the faster and more energetically favorable the attack will be. It's simple electrostatic attraction!

Question

The increasing order of the following compounds towards HCN addition is

Options:

(A) (i) < (iii) < (iv) < (ii)

(B) (iii) < (iv) < (i) < (ii)

(C) (iii) < (iv) < (ii) < (i)

(D) (iii) < (i) < (iv) < (ii)

Sigma Percentile · Accepted Answer

### Mastering Nucleophilic Addition: The Role of Substituent Effects

When dealing with carbonyl compounds like aldehydes and ketones, one of the most fundamental reactions you will encounter is nucleophilic addition. In this problem, we are asked to determine the reactivity order of four substituted benzaldehydes towards the addition of hydrogen cyanide ($HCN$).

### The Core Principle

The addition of $HCN$ involves the attack of the cyanide nucleophile ($CN^-$) on the electrophilic carbonyl carbon. The golden rule here is simple: **the more positive ($\delta+$) the carbonyl carbon, the faster the nucleophilic attack.**

Substituents on the benzene ring can either increase or decrease this positive charge through inductive ($-I$ or $+I$) and resonance ($-R$ or $+R$) effects.

- **Electron Withdrawing Groups (EWG)** pull electron density away from the ring, increasing the $\delta+$ charge on the carbonyl carbon and making the compound **more reactive**.
- **Electron Donating Groups (EDG)** push electron density into the ring, decreasing the $\delta+$ charge and making the compound **less reactive**.

### Analyzing the Competitors

Let's break down the effects of the substituents in our four compounds:

**1. p-Methoxybenzaldehyde (Compound i):**
The $-OCH_3$ group is located at the para position. While it has a weak $-I$ effect, its $+R$ (resonance) effect is incredibly strong. It donates a lone pair of electrons into the aromatic ring, which delocalizes all the way to the carbonyl oxygen. This drastically reduces the electrophilicity of the carbonyl carbon, making it the **least reactive** of the bunch.

**2. p-Nitrobenzaldehyde (Compound ii):**
Here, the $-NO_2$ group is at the para position. The nitro group is a notorious electron-withdrawing powerhouse. It exerts both a $-I$ and a strong $-R$ effect. By pulling electron density out of the ring and away from the carbonyl group, it leaves the carbonyl carbon highly electron-deficient. This makes it the **most reactive** compound.

**3. Benzaldehyde (Compound iii):**
This is our vanilla molecule. With no substituents on the ring, it serves as the perfect **reference point** to compare the others against.

**4. m-Methoxybenzaldehyde (Compound iv):**
This is where many students fall into a trap! The $-OCH_3$ group is at the meta position. **Resonance effects ($+R$ or $-R$) do not operate from the meta position.** Therefore, the methoxy group cannot donate electrons via resonance. It can only exert its $-I$ (inductive) effect, which is electron-withdrawing. Because it withdraws electrons, it makes the carbonyl carbon slightly more positive than in unsubstituted benzaldehyde. Thus, it is **more reactive than benzaldehyde**.

### The Final Verdict

Putting it all together, the $+R$ effect makes compound (i) the least reactive. The $-I$ effect makes compound (iv) more reactive than the reference (iii). Finally, the powerful $-R$ effect makes compound (ii) the undisputed champion of reactivity.

The increasing order of reactivity is:
**(i) < (iii) < (iv) < (ii)**

Animated Solution for Chemistry - Organic Chemistry: The increasing order of the following compounds towards HCN addition is

Select Answer:

Visualized Solution

Nucleophilic Addition of HCN

Role of Substituents

Analyzing Compound (i)

Analyzing Compound (ii)

Analyzing Compound (iii)

Analyzing Compound (iv)

Final Reactivity Order

The Sigma Insight: Carbonyl Compounds

Mastering Nucleophilic Addition

The Core Principle

Analyzing the Competitors

The Final Verdict

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